All categories

3 years ago

5.3 cm 6.6 cm Find the length of the Unknown leg of the right triangle

Mathematics

2 answers:

nika2105 [10]3 years ago

8 0

Answer:

3.93 cm

Step-by-step explanation:

Use the Pythagorean Theorem to solve

$a^{2} + b^{2} = c^{2}$

It is known that the missing value is the leg

Substitute in 5.3 for leg 1 and 6.6 for the hypotenuse

$5.3^{2} + b^{2} = 6.6^{2}$

Solve exponents

$28.09 + b^{2} = 43.56$

Subtract 28.09 from both sides of the equation

$b^{2} = 15.47$

Simplify

b = 3.93 cm

3.93 cm

Hope this helps :)

SSSSS [86.1K]3 years ago

4 0

We are trying to find the unknown leg of the triangle and we can use the formula a^2 + b^2 = c^2.

5.3^2 + b^2 = 6.6^2

28.09 + b^2 = 43.56

28.09 - 28.09 + b^2 = 43.56 - 28.09

b^2 = 15.47

√b^2 = √15.47

b = 3,9331...

Lets round to the bearest hundreth.

3.9331... ≈ 3.93

Therefore, the length of the unknown leg is approximately 3.93cm.

Best of Luck!

You might be interested in

Is (-7,2) a point on the line?

Gala2k [10]

Answer:

what line i swear if u tell me i can answer just comment

Step-by-step explanation:

7 0

3 years ago

Find x. Round your answer to the nearest tenth of a degree.

Kaylis [27]

Answer:

Step-by-step explanation:

hello :

tanx= 15/20

x=36.97°

7 0

3 years ago

Can someone please help me with this algebra question? Image attached.

horsena [70]

Answer:

2 relative maxima and minima

(2, 6.2) = maxima

(0, -4) = minima

*tell me if I got this wrong

3 0

3 years ago

Please help me with number 8

Vika [28.1K]

The answer is negative 3

8 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago